Solve the equation?...and check my answers please.?

n^3 - 9n^2 - n + 9 = 0

When you factorize,the final answer is:

(n-1)(n+1)(n-9)=0

(4x + 9y)^2 =16x^2+72xy+81y^2

-4m^2n^3 - 36m^2n^2=-4m^2n^2(n+9)

9x (-x + 2)=-9x^2+18x

-25x^2 - 115x + 210 = -5(5x-7)(x+6) is correct

n^3 - 9n^2 - n + 9 = 0

factor by grouping

(n^3 - 9n^2) - (n - 9) = 0

n^2 (n - 9) - 1(n - 9) = 0

(n^2 - 1)(n - 9) = 0

(n + 1)(n - 1)(n - 9) = 0

n + 1 = 0 ==> n = -1

n - 1 = 0 ==> n = 1

n - 9 = 0 ==> n = 9

so either n = 1 , -1, or 9

*****

(4x + 9y)^2 = (squaring binomials)

16x^2 + 36xy + 36xy + 81y^2 =

16x^2 + 72xy + 81y^2

*****

-25x^2 - 115x + 210 = -5(5x-7)(x+6)

-25x^2 - 115x + 210 = -5(5x^2 +23x - 42)

-25x^2 - 115x + 210 = -25x^2 - 115x + 210 yep, that's right

*****

GCF: -4m^2n^2, leaving n + 9

you left out a negative sign

-4m^2 n^3 - 36m^2 n^2 = -4m^2 n^2(n + 9)

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left out an x^2:

9x(-x + 2) = -9x^2 + 18x

What are the instructions for the first problem? Is there some lead in material that might help us out. We could use a graphing calulator; but, I don't think that is what you are supposed to do.

If it is the solutions are n = -1, +1, 9.

Squaring the binomial has a rule of sq the first term of the binomial, sq the last term of the binomial, multiply the two pieces of the binomial then multiply by 2.

(4x)^2 +2(4x)(9y) +(9y)^2

16x^2 +72xy +81y^2

It looks like the third problem has already been factored. Good job.

The other two items are also correct. Congratulations.

What is problem